How to show $K_p = \frac{\det II}{\det I}$?

Question

where $K_p$ is the gaussian curvature, $I$ is the first fundamental form and $II$ is the second fundamental form.

Suppose $\xi(u,v)$ is my surface patch. I then computed $II$ as $II_{11} = \langle n_p, \frac{\partial^2 \xi}{ \partial u^2} \rangle $, $II_{12} = II_{21} = \langle n_p, \frac{\partial^2 \xi}{ \partial u \partial v} \rangle $, $II_{22} = \langle n_p, \frac{\partial^2 \xi}{ \partial v^2} \rangle $.

Then, $$\det II = \langle n_p, \frac{\partial^2 \xi}{ \partial u^2} \rangle \cdot \langle n_p, \frac{\partial^2 \xi}{ \partial v^2} \rangle - \left(\langle n_p, \frac{\partial^2 \xi}{ \partial u \partial v} \rangle \right)^2 $$

And I know $$\det I = \left|\left| \frac{\partial \xi}{\partial u} \times \frac{\partial \xi}{\partial v} \right|\right|^2$$

But from here I don't know how to proceed.

Answer 1

Let $\zeta(u,v)$ your surface patch.The gaussian curvature of $\zeta$ is $$K=k_1k_2,$$ where $k_1$ and $k_2$ are the principal curvatures. The first fundamental form can be expressed as $$Edu^2+2Fdudv+G^2dv^2,$$ and the second fundamental form as $$Ldu^2+2Mdudv+Ndv^2.$$ Let \begin{equation}A=\begin{pmatrix}E&F\\ F& G\end{pmatrix},\end{equation} and \begin{equation}B=\begin{pmatrix}L&M\\ M&N\end{pmatrix}.\end{equation} By definition $k_1$ and $k_2$ are the eigenvalues of $A^{-1}B$, so $$K=k_1k_2=\big|A^{-1}B\big |=\frac{LN-M^2}{EG-F^2}$$